Solve 6x^2+x-1

$$6 x ^ { 2 } + x - 1$$

Factor

$$\left(3x-1\right)\left(2x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-1\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite 6x^{2}+x-1 as \left(6x^{2}-2x\right)+\left(3x-1\right).

$$\left(6x^{2}-2x\right)+\left(3x-1\right)$$

Factor out 2x in 6x^{2}-2x.

$$2x\left(3x-1\right)+3x-1$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(2x+1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$6x^{2}+x-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-1\right)}}{2\times 6}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 6\left(-1\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-1±\sqrt{1-24\left(-1\right)}}{2\times 6}$$

Multiply -24 times -1.

$$x=\frac{-1±\sqrt{1+24}}{2\times 6}$$

Add 1 to 24.

$$x=\frac{-1±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-1±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-1±5}{12}$$

Now solve the equation x=\frac{-1±5}{12} when ± is plus. Add -1 to 5.

$$x=\frac{4}{12}$$

Reduce the fraction \frac{4}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-1±5}{12} when ± is minus. Subtract 5 from -1.

$$x=-\frac{6}{12}$$

Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{3} for x_{1} and -\frac{1}{2} for x_{2}.

$$6x^{2}+x-1=6\left(x-\frac{1}{3}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$6x^{2}+x-1=6\left(x-\frac{1}{3}\right)\left(x+\frac{1}{2}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+x-1=6\times \frac{3x-1}{3}\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+x-1=6\times \frac{3x-1}{3}\times \frac{2x+1}{2}$$

Multiply \frac{3x-1}{3} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+x-1=6\times \frac{\left(3x-1\right)\left(2x+1\right)}{3\times 2}$$

Multiply 3 times 2.

$$6x^{2}+x-1=6\times \frac{\left(3x-1\right)\left(2x+1\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}+x-1=\left(3x-1\right)\left(2x+1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 6

$$x ^ 2 +\frac{1}{6}x -\frac{1}{6} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{1}{6} rs = -\frac{1}{6}$$

Two numbers r and s sum up to -\frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{6} = -\frac{1}{12}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{1}{12} - u s = -\frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{6}

$$(-\frac{1}{12} - u) (-\frac{1}{12} + u) = -\frac{1}{6}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1}{144} - u^2 = -\frac{1}{6}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{1}{6}-\frac{1}{144} = -\frac{25}{144}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{144} u = \pm\sqrt{\frac{25}{144}} = \pm \frac{5}{12}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{1}{12} - \frac{5}{12} = -0.500 s = -\frac{1}{12} + \frac{5}{12} = 0.333$$

Evaluate

$$6x^{2}+x-1$$